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lbvjy [14]
1 year ago
10

23.It is possible to dilute a solution by...Select one:a. adding more solute to the solution.b. adding more solvent to the solut

ion.c. lowering the temperature of the solution.d. stirring the solution.
Chemistry
1 answer:
joja [24]1 year ago
7 0

To dilute a solute in a solution, it is necessary to add a proper solvent for the reaction to occur, and adding more solvent will cause the solution to dilute even more, therefore the best answer will be letter B

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The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W
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Answer:

0.0010 mol·L⁻¹s⁻¹  

Explanation:

Assume the rate law is  

rate = k[A][B]²

If you are comparing two rates,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}

You are cutting each concentration in half, so

\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}

Then,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}

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3 years ago
Draw the resonance structures for the conjugate base of Phenol (C6H6O). In one sentence, explain why phenol (C6H6O) has a pKa of
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Answer:

The four resonance structures of the phenoxide ion are shown in the image attached

The conjugate base of cyclohexanol has only one resonance contributor, while

the conjugate base of phenol has four resonance contributors.

Explanation:

In organic chemistry, it is known that structures are more stable if they possess more resonance contributors. The greater the number of contributing canonical structures, the more stable the organic specie. Since the phenoxide ion has four contributing canonical structures, it is quite much more stable than cyclohexanol having only one contributing structure to its conjugate base. Hence the PKa(acid dissociation constant) of phenol is lesser than that of cyclohexanol. The conjugate base of phenol is stabilized by resonance.

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